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#### A massless string connects two pulley of masses ' $2 \mathrm{~kg}$' and '$1 \mathrm{~kg}$' respectively as shown in the figure.The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, $g=10 \mathrm{~m} / \mathrm{s}^2$]Option: 1 $\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2$Option: 2 $\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2$Option: 3 $\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2$Option: 4 $\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2$

Not understanding sir

1.32

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#### A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure.  The coefficient of friction, between the particle and the rough track equals µ.  The particle is released, from rest, from the point P and it comes to rest at a point R.  The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(=QR), are, respectively close to : Option: 1  0.2 and 6.5 m   Option: 3  0.2 and 3.5 m   Option: 4   0.29 and 6.5 m

Work done by friction at QR = μmgx

In triangle, sin 30° = 1/2 = 2/PQ

PQ = 4 m

Work done by friction at PQ = μmg × cos 30° × 4 = μmg × √3/2 × 4 = 2√3μmg

Since work done by friction on parts PQ and QR are equal,

μmgx = 2√3μmg

x = 2√3 ≅ 3.5 m

Applying work energy theorem from P to R

decrease in P.E.=P.E.= loss of energy due to friction in PQPQ and QR

$\\ m g h=(\mu m g \cos \theta) P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu \times Q R =\mu \cos 30^{\circ} \times 4+\mu \times 2 \sqrt{3} =\mu\left(4 \times \frac{\sqrt{3}}{2}+2 \sqrt{3}\right)\\ h=\mu \times 4 \sqrt{3}\\ \mu=\frac{2}{4 \sqrt{3}}=\frac{1}{2 \sqrt{3}}=0.29$where h=2(given)

#### A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed in the x-y plane as shown in the figure : Which of the following statements is false for the angular momentum  about the origin? Option: 1 $\vec{L}= -\frac{mv}{\sqrt{2}}R\hat{k}$when the particle is moving from A to B. Option: 2 $\vec{L}= mv\left [ \frac{R}{\sqrt{2}}-a \right ]\hat{k}$when the particle is moving from C to D. Option: 3 $\vec{L}= mv\left [ \frac{R}{\sqrt{2}}+a \right ]\hat{k}$when the particle is moving from B to C. Option: 4 when the particle is moving from D to A.

$\\ In\ option\ (a), co-ordinates \ of\ A are \left(\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}\right) \\ \therefore \vec{r}=\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) and \vec{v}=v \hat{i}\\ \vec{L} m(\vec{r} \times \vec{v})=m\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) \times v \hat{i}\\ \vec{L}=-\frac{m R}{\sqrt{2}} v \hat{k}$

$\\ in\ option \ (b)\ it \ moves\ from \ C \ to\ D\\ L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})$

so option b is correct option

$\\ in \ option\ (c), \ For\ B \ to\ C, \\ we have L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})$

$\\ in \ option \ (d), \ When \ a \ particle\ is \ moving\ from \ D \ to\ A\\ L=\frac{R}{\sqrt{2}} m v(-k)$

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#### Consider a uniform rod of mass$M=4m$ and length l pivoted about its centre. A mass m moving with velocity $v$ making angle $\theta =\frac{\pi }{4}$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod - mass system just after the collision is :    Option: 1 Option: 2  Option: 3    Option: 4

Let us conserve angular momentum about O:-

So, $L_i=\left ( \frac{mv}{\sqrt2} \right )\times \frac{l}{2}$, where $\left ( \frac{mv}{\sqrt2} \right )$ is linear momentum and $\left ( \frac{l}{2} \right )$ is the distance from centre O.

Now, $L_f=I\omega$

Here, $I=\frac{4ml^2}{12}+{m\left (\frac{l}{2} \right )^2}=\frac{7ml^2}{12}$

So, $L_i=L_f\Rightarrow \omega=\frac{6v}{7\sqrt2 l}= \frac{3\sqrt2v}{7 l}$

So the correct graph is given in option 2.

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#### A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed  about the fixed end of the spring such that it rotates in a circle in gravity-free space. Then the stretch in the spring is :      Option: 1   Option: 2  Option: 3    Option: 4

As natural lentgh=l

Let elongation=x

Mass m is moving with angular velocity  in a radius r

where

Due to elongation x spring force is given by

And

as

So

So the correct option is 2.

#### The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass . (The coordinates of the same are shown in the figure) are : Option: 1 Option: 2  Option: 3  Option: 4

The co-ordinate of O1 is (0.5,1),  O2 is (1, 2.5)

So the correct option is 3.

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#### A spring mass system (mass m, spring constant k and natural length l ) rests in equilibrium on a horizontal disc.The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about its axis with an angular velocity $\omega$, $(k> > m\omega ^{2})$ the relative change in the length of the spring is best given by the option : Option: 1       Option: 2          Option: 3         Option: 4

As natural lentgh=l0

Let elongation=x

Mass m is moving with angular velocity $\omega$ in a radius r

where $r=l_{0}+x$

Due to elongation x spring force is given by $F_{s}=Kx$

And $F_{C}=m\omega ^{2}r=m\omega ^{2}(l_{0}+x)$

as $F_{C}=F_{s}$

$Kx=m\omega ^{2}(l_{0}+x)$

$\Rightarrow x=\frac{m\omega ^{2}l_{0}}{K-m\omega ^{2}}$

Using $k> > m\omega ^{2}$

So, $\frac{x}{l_{0}}$ is equal to $\frac{m\omega ^{2}}{k}$

Hence the correct option is (2).

#### Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm, and 5.0 cm as shown in the figure. The centre of mass of the system is at a point:  Option: 1 right and above  mass     Option: 2  right and  above  mass Option: 3  right and  above  mass   Option: 4  right and  above  mass

let m1=1 kg, m2=1.5 kg and m3=2.5 kg

x1=0, x2=3, x3=0 and y1=0, y2=0, y3=4

and

Let point A be origin and mass m1=1.0 kg be at origin.

So,

and

so centre of mass of the system is at (0.9,2).

So from the figure we can say that the 0.9 cm right and 2 cm above the 1 kg mass.

So option (2) is correct.